Question

Find the position and the nature of the image formed when an object is placed at a distance of 5 cm from a
concave lens of focal length 15 cm. What is its magnification?​

Answer 1

Given:

The distance of the object(u) = 5 cm.

Focal length (f) = 15 cm.

To find:

• The position of the image.
• The nature of the image.
• The magnificent of the image.

Solution:

Let, the distance of the image from the lens be, 'v'.

The magnificent of the image be, 'm'.

The size of the image be, 'I'.

f= 15 cm

u = 5 cm

From the lens formula we get that,

$\dfrac{1}{f} = (\dfrac{1}{u} - \dfrac{1}{v})$

$\\ or, \dfrac{1}{15} = (\dfrac{1}{5}-\dfrac{1}{v})$

$or, (\dfrac{1}{5} - \dfrac{1}{v}) = \dfrac{1}{15}$

$or, (-\dfrac{1}{v})= ( \dfrac{1}{15} - \dfrac{1}{5})$

$or, (-\dfrac{1}{v}) = (\dfrac{1-3}{15})$

$or, (-\dfrac{1}{v})= \dfrac{-2}{15}$

$or, \dfrac{1}{v} = \dfrac{2}{15}$

$or, v = \dfrac{15}{2}$

or, v = 7.5 cm.

∴ The image has formed at a distance of 7.5 cm in front of the lens.

From the formula we get that,

$m = \dfrac{-v}{u}$

$or, m = \dfrac{-(7.5)}{5}$

or, m = -1.5 [Magnificent is a ratio so, it has no unit.]

∴The magnificent  of the image is  -1.5.

Since, the magnificent of the image is negative so, the image is real, inverted,  and smaller in size.

Answer:

The image has formed 7.5 cm in front of the mirror.              

The image is real, inverted and smaller in size.

The magnificent of the image is -1.5.

Answer 2

HOPE IT IS HELPFUL TO YOU.